matrix homomorphism












-1












$begingroup$


Prove that $ϕ$ is a homomorphism and describe its kernel.
$$
ϕ:ℝ→GL(2,ℝ),qquad
ϕ(x)=
begin{bmatrix}cos(2x)&sin(2x)\-sin(2x)&cos(2x)end{bmatrix}
$$
where $x∈ℝ$



I have begun by saying
$$
y=begin{bmatrix}cos(2y)&sin(2y)\-sin(2y)&cos(2y)end{bmatrix}
$$
$ℝ$ denote the set of real numbers under addition so I need to show that $ϕ(x+y)=ϕ(x)ϕ(y)$



So I have calculated
$$
ϕ(x+y)begin{bmatrix}cos(2y)+cos(2y)&sin(2y)+sin(2y)\-sin(2y)-sin(2y)&cos(2y)+cos(2y)end{bmatrix}
$$
Is this right? And I don't know how to go from there. any help would be appreciated.






abstract-algebra group-homomorphism







share|cite|improve this question











$endgroup$

















    -1












    $begingroup$


    Prove that $ϕ$ is a homomorphism and describe its kernel.
    $$
    ϕ:ℝ→GL(2,ℝ),qquad
    ϕ(x)=
    begin{bmatrix}cos(2x)&sin(2x)\-sin(2x)&cos(2x)end{bmatrix}
    $$
    where $x∈ℝ$



    I have begun by saying
    $$
    y=begin{bmatrix}cos(2y)&sin(2y)\-sin(2y)&cos(2y)end{bmatrix}
    $$
    $ℝ$ denote the set of real numbers under addition so I need to show that $ϕ(x+y)=ϕ(x)ϕ(y)$



    So I have calculated
    $$
    ϕ(x+y)begin{bmatrix}cos(2y)+cos(2y)&sin(2y)+sin(2y)\-sin(2y)-sin(2y)&cos(2y)+cos(2y)end{bmatrix}
    $$
    Is this right? And I don't know how to go from there. any help would be appreciated.






    abstract-algebra group-homomorphism







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      Prove that $ϕ$ is a homomorphism and describe its kernel.
      $$
      ϕ:ℝ→GL(2,ℝ),qquad
      ϕ(x)=
      begin{bmatrix}cos(2x)&sin(2x)\-sin(2x)&cos(2x)end{bmatrix}
      $$
      where $x∈ℝ$



      I have begun by saying
      $$
      y=begin{bmatrix}cos(2y)&sin(2y)\-sin(2y)&cos(2y)end{bmatrix}
      $$
      $ℝ$ denote the set of real numbers under addition so I need to show that $ϕ(x+y)=ϕ(x)ϕ(y)$



      So I have calculated
      $$
      ϕ(x+y)begin{bmatrix}cos(2y)+cos(2y)&sin(2y)+sin(2y)\-sin(2y)-sin(2y)&cos(2y)+cos(2y)end{bmatrix}
      $$
      Is this right? And I don't know how to go from there. any help would be appreciated.






      abstract-algebra group-homomorphism







      share|cite|improve this question











      $endgroup$




      Prove that $ϕ$ is a homomorphism and describe its kernel.
      $$
      ϕ:ℝ→GL(2,ℝ),qquad
      ϕ(x)=
      begin{bmatrix}cos(2x)&sin(2x)\-sin(2x)&cos(2x)end{bmatrix}
      $$
      where $x∈ℝ$



      I have begun by saying
      $$
      y=begin{bmatrix}cos(2y)&sin(2y)\-sin(2y)&cos(2y)end{bmatrix}
      $$
      $ℝ$ denote the set of real numbers under addition so I need to show that $ϕ(x+y)=ϕ(x)ϕ(y)$



      So I have calculated
      $$
      ϕ(x+y)begin{bmatrix}cos(2y)+cos(2y)&sin(2y)+sin(2y)\-sin(2y)-sin(2y)&cos(2y)+cos(2y)end{bmatrix}
      $$
      Is this right? And I don't know how to go from there. any help would be appreciated.






      abstract-algebra group-homomorphism




      abstract-algebra group-homomorphism






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 22:11









      egreg

      183k1486204




      183k1486204










      asked Dec 3 '18 at 21:56









      jessicajessica

      42




      42






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          Note that by definition,
          $$
          phi(x+y)=
          begin{bmatrix}
          cos(2(x+y)) & sin(2(x+y)) \
          -sin(2(x+y)) & cos(2(x+y))
          end{bmatrix}
          $$
          Now note that
          begin{align}
          cos(2x+2y)&=cos(2x)cos(2y)-sin(2x)sin(2y) \
          sin(2x+2y)&=sin(2x)cos(2y)+cos(2x)sin(2y)
          end{align}
          and compute
          $$
          phi(x)phi(y)=
          begin{bmatrix}
          cos(2x) & sin(2x) \
          -sin(2x) & cos(2x)
          end{bmatrix}
          begin{bmatrix}
          cos(2y) & sin(2y) \
          -sin(2y) & cos(2y)
          end{bmatrix}
          $$
          as a matrix product.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
            $endgroup$
            – jessica
            Dec 4 '18 at 11:50












          • $begingroup$
            @jessica Why the determinant?
            $endgroup$
            – egreg
            Dec 4 '18 at 12:37










          • $begingroup$
            :Don't worry I've found it. But can you help me with finding the kernel?
            $endgroup$
            – jessica
            Dec 4 '18 at 12:54












          • $begingroup$
            @jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
            $endgroup$
            – egreg
            Dec 4 '18 at 13:02













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          1 Answer
          1






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          1 Answer
          1






          active

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          active

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          active

          oldest

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          0












          $begingroup$

          Note that by definition,
          $$
          phi(x+y)=
          begin{bmatrix}
          cos(2(x+y)) & sin(2(x+y)) \
          -sin(2(x+y)) & cos(2(x+y))
          end{bmatrix}
          $$
          Now note that
          begin{align}
          cos(2x+2y)&=cos(2x)cos(2y)-sin(2x)sin(2y) \
          sin(2x+2y)&=sin(2x)cos(2y)+cos(2x)sin(2y)
          end{align}
          and compute
          $$
          phi(x)phi(y)=
          begin{bmatrix}
          cos(2x) & sin(2x) \
          -sin(2x) & cos(2x)
          end{bmatrix}
          begin{bmatrix}
          cos(2y) & sin(2y) \
          -sin(2y) & cos(2y)
          end{bmatrix}
          $$
          as a matrix product.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
            $endgroup$
            – jessica
            Dec 4 '18 at 11:50












          • $begingroup$
            @jessica Why the determinant?
            $endgroup$
            – egreg
            Dec 4 '18 at 12:37










          • $begingroup$
            :Don't worry I've found it. But can you help me with finding the kernel?
            $endgroup$
            – jessica
            Dec 4 '18 at 12:54












          • $begingroup$
            @jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
            $endgroup$
            – egreg
            Dec 4 '18 at 13:02


















          0












          $begingroup$

          Note that by definition,
          $$
          phi(x+y)=
          begin{bmatrix}
          cos(2(x+y)) & sin(2(x+y)) \
          -sin(2(x+y)) & cos(2(x+y))
          end{bmatrix}
          $$
          Now note that
          begin{align}
          cos(2x+2y)&=cos(2x)cos(2y)-sin(2x)sin(2y) \
          sin(2x+2y)&=sin(2x)cos(2y)+cos(2x)sin(2y)
          end{align}
          and compute
          $$
          phi(x)phi(y)=
          begin{bmatrix}
          cos(2x) & sin(2x) \
          -sin(2x) & cos(2x)
          end{bmatrix}
          begin{bmatrix}
          cos(2y) & sin(2y) \
          -sin(2y) & cos(2y)
          end{bmatrix}
          $$
          as a matrix product.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
            $endgroup$
            – jessica
            Dec 4 '18 at 11:50












          • $begingroup$
            @jessica Why the determinant?
            $endgroup$
            – egreg
            Dec 4 '18 at 12:37










          • $begingroup$
            :Don't worry I've found it. But can you help me with finding the kernel?
            $endgroup$
            – jessica
            Dec 4 '18 at 12:54












          • $begingroup$
            @jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
            $endgroup$
            – egreg
            Dec 4 '18 at 13:02
















          0












          0








          0





          $begingroup$

          Note that by definition,
          $$
          phi(x+y)=
          begin{bmatrix}
          cos(2(x+y)) & sin(2(x+y)) \
          -sin(2(x+y)) & cos(2(x+y))
          end{bmatrix}
          $$
          Now note that
          begin{align}
          cos(2x+2y)&=cos(2x)cos(2y)-sin(2x)sin(2y) \
          sin(2x+2y)&=sin(2x)cos(2y)+cos(2x)sin(2y)
          end{align}
          and compute
          $$
          phi(x)phi(y)=
          begin{bmatrix}
          cos(2x) & sin(2x) \
          -sin(2x) & cos(2x)
          end{bmatrix}
          begin{bmatrix}
          cos(2y) & sin(2y) \
          -sin(2y) & cos(2y)
          end{bmatrix}
          $$
          as a matrix product.






          share|cite|improve this answer











          $endgroup$



          Note that by definition,
          $$
          phi(x+y)=
          begin{bmatrix}
          cos(2(x+y)) & sin(2(x+y)) \
          -sin(2(x+y)) & cos(2(x+y))
          end{bmatrix}
          $$
          Now note that
          begin{align}
          cos(2x+2y)&=cos(2x)cos(2y)-sin(2x)sin(2y) \
          sin(2x+2y)&=sin(2x)cos(2y)+cos(2x)sin(2y)
          end{align}
          and compute
          $$
          phi(x)phi(y)=
          begin{bmatrix}
          cos(2x) & sin(2x) \
          -sin(2x) & cos(2x)
          end{bmatrix}
          begin{bmatrix}
          cos(2y) & sin(2y) \
          -sin(2y) & cos(2y)
          end{bmatrix}
          $$
          as a matrix product.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '18 at 12:41

























          answered Dec 3 '18 at 22:14









          egregegreg

          183k1486204




          183k1486204












          • $begingroup$
            @ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
            $endgroup$
            – jessica
            Dec 4 '18 at 11:50












          • $begingroup$
            @jessica Why the determinant?
            $endgroup$
            – egreg
            Dec 4 '18 at 12:37










          • $begingroup$
            :Don't worry I've found it. But can you help me with finding the kernel?
            $endgroup$
            – jessica
            Dec 4 '18 at 12:54












          • $begingroup$
            @jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
            $endgroup$
            – egreg
            Dec 4 '18 at 13:02




















          • $begingroup$
            @ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
            $endgroup$
            – jessica
            Dec 4 '18 at 11:50












          • $begingroup$
            @jessica Why the determinant?
            $endgroup$
            – egreg
            Dec 4 '18 at 12:37










          • $begingroup$
            :Don't worry I've found it. But can you help me with finding the kernel?
            $endgroup$
            – jessica
            Dec 4 '18 at 12:54












          • $begingroup$
            @jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
            $endgroup$
            – egreg
            Dec 4 '18 at 13:02


















          $begingroup$
          @ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
          $endgroup$
          – jessica
          Dec 4 '18 at 11:50






          $begingroup$
          @ egreg: do I need to calculate the determinant of each and then multiply together? if so I did it but didn't get the same answer.
          $endgroup$
          – jessica
          Dec 4 '18 at 11:50














          $begingroup$
          @jessica Why the determinant?
          $endgroup$
          – egreg
          Dec 4 '18 at 12:37




          $begingroup$
          @jessica Why the determinant?
          $endgroup$
          – egreg
          Dec 4 '18 at 12:37












          $begingroup$
          :Don't worry I've found it. But can you help me with finding the kernel?
          $endgroup$
          – jessica
          Dec 4 '18 at 12:54






          $begingroup$
          :Don't worry I've found it. But can you help me with finding the kernel?
          $endgroup$
          – jessica
          Dec 4 '18 at 12:54














          $begingroup$
          @jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
          $endgroup$
          – egreg
          Dec 4 '18 at 13:02






          $begingroup$
          @jessica The condition for $xinkerphi$ is $cos(2x)=1$ and $sin(2x)=0$, so $phi(x)$ is the identity matrix.
          $endgroup$
          – egreg
          Dec 4 '18 at 13:02




















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